What's in the box?

This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?

Age

7 to 11

Challenge level

Primary

Number - Multiplication, Division and Ratio

Problem Solving

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

What's in the Box? printable sheet

In the picture below, four whole numbers are being put into the box. Inside the box, a multiplication happens to each number, and then four new numbers are tipped out of the box: 56, 24, 112 and 216.

What multiplication might have happened to each number inside the box to get the answers in the picture above?

Are there any other possibilities?

What's the largest number that each of the four starting numbers might have been multiplied by inside the box? How do you know?

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Imagine that four more numbers are put into the box, but now the box multiplies all of them by a new number. The numbers that come out of the box are:

143

297

341

1221

What number might the box be multiplying by? How do you know?

Discuss this with some other people and see if there are any different ways to work this out.

Student Solutions

We had a good number of solutions sent in with explanations as to how the answers were found. Ansh from Monkfield Park Primary School sent in the following:

For question 1) Answer is 8.

First of all I thought that 7 times 8 will be 56. Then I tried 3 times 8 and that also worked. Next I tried 13 times 8 that was 98 but that did not work so I tried 14 times 8 which was 112 and it was correct. Finally I divided 216 by 8 and ended up with 7.

So that's how I figured out that the multiplier is 8.

For question 2) Answer is 11

First of all I tried 3 and that did not work so I tried 6 and that did not work either. I also tried 7 and 9 but they both were wrong too. I knew it was not an even number so I skipped 2 and 10. Then I tried 11 and that worked.

So the first one was 13 times 11 = 143.

Next I divided 297 by 11 and the answer was 297.

After that I tried 341 divided by 11 and came up with 31.

Finally I did 1221 divided by 11 which gave me 111.

So the multiplier was 11.

Delaney at Mount Vernon School Maine, U.S. said:

Question 1 Possible factors: 2, 4 or 8

Largest common factor: 8

$56\div2=28$, $28\div2=14$, $14\div2=7$

$24\div2=12$, $12\div2=6$, $6\div2=3$

$112\div2=56$, $56\div2=28$, $28\div2=14$

$216\div2=108$, $108\div2=54$, $54\div2=27$

After I got these answers there were no more common factors and I assumed that 7, 3, 14, and 27 were the original numbers. 8 is the largest common factor because $2^3$ or 2x2x2=8.

Question 2 Only possible factor: 11

None of the numbers were even in this question so I knew that no multiples of two would be a common factor.

143=110+33

297=220+77

1221=1100+121

341=330+11

After dividing into portions divisible by 11, I found the original numbers by dividing by 11.

$143\div11=13$

$341\div11=31$

$1221\div11=111$

$297\div11=27$

There were no more common factors, so I assumed those were the original numbers.

We had this submitted from Jerry at Dulwich College, Shanghai:

First I checked which numbers that 143 can be divided by.

I skipped the multiples of 5 because 143, 297, 341 and 1221 clearly wouldn't be able to be divided by multiples of 5. I first found out 143 can be divided by 11 and 13. But 297 cannot be divided by 13. So I kept working on 11 and all four numbers can be divided by 11. So the winner is 11!

Well, altogether these have been very good responses to this challenge. Keep sending solutions to any other activity you have a go at.

Teachers' Resources

What's in the Box?

Four numbers in little boxes are put into a special big box that does a multiplication, then four new numbers come out at the end:

We only used whole numbers to go in, so, what multiplication might have gone on in the big box to get the answers in the picture above?

What was the largest number that could have been used to multiply by, in that big box?

Imagine four new boxes now (with new numbers in) and the large box multiplying by a different number this time. The numbers that come out are these:

What would be the number that the big box is multiplying by?

How are you working these out?

Discuss with others and see if there are different ways that you found the answers.

Why do this problem?

Of course this problem is rather like a function machine, but it can be more interesting for the pupils and easily extended to challenge a wide range of pupils. It could be used to introduce children to the idea of common factors and offers opportunities for learners to record in whatever way they choose.

Possible approach

It may be necessary to introduce the class to just one number going in and to give them one outcome to start with so that they understand the process. Then, gradually increase the number of numbers going in until you reach four, as in the problem. Your own examples can be adjusted in complexity according to the level of your pupils.

Once learners have had some time to work on the first part of the problem in pairs, ask them to share their ways of working with the whole group. Look out for those who give good reasons for choosing particular methods. At this stage, you could introduce the vocabulary of common factors if appropriate.

You may also wish to draw attention to interesting ways of recording. Some children may have drawn pictures, others may have written calculations and others may have done both.

Key questions

What might have gone on in the box to get this number answer?

Could that have produced the other answers too?

Possible extension

Pupils can be challenged to explain how they know they've found the largest possible number that the inputs have been multiplied by. Some pupils could go on to invent their own similar problems for others to do, including ones where a division is performed inside the box instead of a multiplication.

Possible support

Some pupils might benefit from beginning with the version of this task that uses addition and subtraction: What Was in the Box?

In order to introduce the multiplication version of this task, you can use sticks of cubes (such as multilink cubes) to represent one number going into the 'box'. Cover the cubes with a cloth and then secretly add the required extra number of cubes under the cover, in the form of more sticks of cubes, before revealing them to the child. Then a number of probing questions can be asked: How many cubes are there now? What has happened under the cover? Encourage pupils to consider how the number of sticks has grown, e.g. from 1 to 5, instead of simply counting how many more cubes there are.

As they tackle the main problem, some learners might find it useful to have a multiplication square or calculator available.

Money bags

Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?

Age

5 to 11

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

Ram divided $15$ pennies among four small bags.

He labelled each bag with the number of pennies inside it.

He could then pay any sum of money from $1$p to $15$p without opening any bag.

How many pennies did Ram put in each bag?

This problem is based on Money Bags from 'Mathematical Challenges for Able Pupils Key Stages 1 and 2', published by DfES. You can download a copy here.

Student Solutions

We had quite a few submissions for this challenge and these make interesting reading.

Harrison School sent in a solution from Charlie and Oscar as follows;

To begin, we decided to use post-it-notes as bags and sweets to represent our pennies - this made it easy to visualise the question. Then, we worked out that one could only be added to zero in order to make itself; there was now a bag with one penny in it. After that, we made a bag with two pennies in. We did think that we could make two with two ones yet that would give us a problem later. Next, we worked out that three could already be made using bags one and two. However, four could not be made with bags one and two, therefore we had to make bag three have four pennies in it. We then looked and saw we could make numbers five (bag one and three), six (bags three and two) and seven (bags one, two and three) but eight was unable to be made. Consequently, that means that eight has to be in bag four. This used up all the pennies - there were none left over. Finally, we checked that the numbers nine to fifteen worked, which they did. Nine = bags four and one. Ten = bags four and two. Eleven = bags four, two and one. Twelve = bags four and three. Thirteen = bags four, three and one. Fourteen = bags four, three and two. Fifteen = bags four, three, two and one.

Uma from Westhill Primary School, Aberdeenshire sent in her thoughts and the solution.

The solution for the Money Bags problem is 1,2,4 and 8 Here is the explanation of how I got my answer:

There are four bags so there can only be four different types of coins. The basic idea behind the solution is to find the four different numbers which can add up to give 1p to 15p. The solution for the question is 1,2,4 and 8. The 1st bag holds 1x1p, the 2nd bag contains 2x1p, the 3rd bag holds 4x1p and the 4th bag contains 8x1p. We can get any number between 1 to 15 using any combination of 1,2,4 and 8.

1p = Bag 1; 2p = Bag 2; 3p = Bag 1 + Bag 2; 4p = Bag 3; 5p = Bag 3 + Bag 1

6p = Bag 3 + Bag 2; 7p = Bag 3 + Bag 2 + Bag 1; 8p = Bag 4; 9p = Bag 4 + Bag 1

10p = Bag 4 + Bag 2; 11p = Bag 4 + Bag 2 + Bag 1; 12p = Bag 4 + Bag 3;

13p = Bag 4 + Bag 3 + Bag 1; 14p = Bag 4 + Bag 3 + Bag 2

15p = Bag 4 + Bag 3 + Bag 2 + Bag 1

Mr Howes' Magnificent Maths Group from Pierrepont Gamston Primary School

As part of an intervention group, we were a little confused to start off with - we didn't realise that the question was asking us to split the money into four bags so that the bags' totals could be combined to make any value up to 15p. Once we worked this out we wrote the values 1p to 15p down the side of the page and tried to find a system to solving the problem.

1p = this was easy, it had to be made with a bag with 1p in it!

2p = this could be made by another bag having 1p in it, or a second bag having 2p in it. We went for the latter because it stopped us using up our bags too quickly!

3p = this was also easy: 1p+2p

4p = We could either have another 1p or 2p bag, or a new 4p bag. We used the 4p bag.

At this point we spotted a pattern - 1p, 2p, 4p... would the next bag contain 8p?

5p = 4p+1p; 6p = 4p+2p; 7p = 4p+2p+1p; 8p = We stuck to our prediction and put 8p in the final bag!; 9p = 8p+1p; 10p = 8p+2p; 11p = 8p+2p+1p; 12p = 8p +4p; 13p = 8p+4p+1p

14p = 8p+4p+2p; 15p = 8p+4p+2p+1p

We thought that if the pattern was continued, the next bag should contain 16p ( making the total value 29p).

We wondered if this would be a better system for the values of coins; however, we realised that the values we used help us add up the numbers mentally much faster and make finding change easier!

James, Elliot, Freddie, Anna, Orla and Oliver

Jake from Duke of Norfolk School said

We drew out circles to show the different bags.

We tried to make all the numbers from 1 to 15, and recorded the number of pennies used in the circles.

We started with 1, 2, 3 and 5 coins, but found that we could not make numbers up from 12. We realised that we had to keep 1 and 2, because they were needed to pay 1p and 2p. So we added coins to the other bags, eventually settling on 1, 2, 4 and 8.

Stuart from Dussindale Primary School sent in this work and a good picture illustrating the solution

I started to look at multiples of ten but then realised I needed to use doubles/halves.

My first pair were 2 and 8 and then I saw I needed to have 1 and 2 and 4 and 8.

I used the cubes to make sure that I could make all the numbers from 1 to 15.

Sarah from Cabot Primary School, Bristol sent in the following;

A Y4 maths group from Cabot Primary School, Bristol (Iman, Andrea-Marie, Sakariya and Aisha) worked on this problem together.

First of all, we found 15 play coins and a painting tray to help us think about the investigation and find a solution to the maths problem.

Secondly, we decided that we definitely needed to have one bag with 1p in it (or we would not be able to pay for something that cost 1p!), so we put 1p in one of the pots.

Then we thought about which numbers could make all of the other prices up to 15p. We realised that if we also had a bag with 2p in it, we could make 2p and 3p (1p + 2p), so we put 2p in our second pot. We tried 3p in the third pot and this gave us more prices:

1p (1p); 2p (2p); 3p (3p); 4p (3p + 1p); 5p (3p + 2p); 6p (3p + 2p + 1p)

...but we couldn't make 7p.

So we tried again, this time with 4p in the third pot. We could then make all of the prices up to 7p.

1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).

We only had one empty pot left, so we put all of the rest of the coins into it - there were 8 of them. We then found that you could make all of the other numbers:

8p (8p); 9p (8p + 1p); 10p (8p + 2p); 11p (8p + 2p + 1p); 12p (8p + 4p); 13p (8p + 4p + 1p);   14p (8p + 4p + 2p); 15p (8p + 4p + 2p + 1p)

So we knew we had found the solution! We tried all the possibilities and didn't give up because Cabot Primary children are resilient!  PS Our paint tray actually has six pots in it. We can see a pattern building up and are now going to think about how to investigate the maximum number of prices we can make if we use all six pots.

Later Sarah wrote and sent in this;

When we looked again at the number of coins in each pot, we noticed that each number was double the previous one. Since double 8 is 16, we put 16 coins into the next empty pot so we had five pots containing 1p, 2p, 4p, 8p and 16p. We knew the maximum price we could now pay would be 31p (our existing 15p plus the extra pot of 16p). We checked that we could now make all of the prices from 15p to 31p (which we could - by adding the new pot of money to each of the combinations that we had already found). We knew that we had found a system and that the sixth pot should therefore contain 32p (which is double 16p) and that we would be able to make all of the prices up to a maximum of 63p.

We then came up with this sequence to show how much money twelve pots (two paint trays) would contain:

1p,    2p,    4p,    8p,    16p,    32p,    64p,    £1.28,    £2.56,    £5.12,     £10.24     £20.48

It was good to hear about the different ways that solutions were found. Well done all of you.

Teachers' Resources

Money Bags

Ram divided $15$ pennies among four small bags.

He labelled each bag with the number of pennies inside it.

He could then pay any sum of money from $1$p to $15$p without opening any bag.

How many pennies did Ram put in each bag?

This problem is based on Money Bags from 'Mathematical Challenges for Able Pupils Key Stages 1 and 2', published by DfES. You can find out more about this book, including how to order it, on the Standards website here .